Question

A ball of mass m moves with speed v and strikes a wall having infinity mass and it turns with the same speed then, find the work done by the ball on the wall.
A. Zero
B. mv J
C. \[\dfrac{m}{v}J\]
D. \[\dfrac{v}{m}J\]

Answer 1

Hint:Before going to solve this question we need to understand the work done and the work-energy theorem. Work is nothing but a force needed to move an object from one place to another. The work-energy theorem states that the work done by the force is stored in a system in the form of energy. Work done can be positive, negative, or zero depending on the angle between the force applied and the displacement.

Formula Used:
To find the work done we have,
\[W = \Delta E\]
Where \[\Delta E\] is change in kinetic energy.

Complete step by step solution:
Consider a ball of mass m moving with a speed v and strikes a wall having infinite mass and the velocity is zero. When the ball hits the wall it turns with the same speed v then, we need to find the work done by the ball on the wall. By the formula of work done, we know that,
Work done = change in kinetic energy
\[W = \Delta E\]……… (1)
\[\Rightarrow \Delta E = {E_2} - {E_1} \\ \]
\[\Rightarrow {E_1} = \dfrac{1}{2}m{v_1}^2 \\ \]
\[\Rightarrow {E_2} = \dfrac{1}{2}m{v_2}^2 \\ \]
By substituting the values of \[{E_1} = \dfrac{1}{2}m{v_1}^2\] and \[{E_2} = \dfrac{1}{2}m{v_2}^2\]in equation (1) we get,
\[W = \dfrac{1}{2}m{v_2}^2 - \dfrac{1}{2}m{v_1}^2 \\ \]
Hence, when the ball hits the wall, it returns with the same velocity that is \[{v_1} = {v_2} = v\] then,
\[W = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{v^2}\]
\[\therefore W = 0\]
Therefore, when the work done by the ball on the wall is zero.

Hence, option A is the correct answer.

Note:We can say that, the work done is zero because the Work done on an object depends on the amount of force F causing the work, the displacement d that experienced by the object.